Question

In a Young's double slit experiment, the slits are $2 \,mm$ apart and are illuminated with a mixture of two wavelength $\lambda_{0}=750 \,nm$ and $\lambda=900 \,nm$. The minimum distance from the common central bright fringe on a screen $2 \,m$ from the slits where a bright fringe from one interference pattern coincides with a bright fringe from the other is

• A. 1.5 mm
• B. 3 mm
• C. 4.5 mm
• D. 6 mm

Answer 1

Answer: (C)

From the given data, note that the fringe width $\left(\beta_{1}\right)$ for $\lambda_{1}=900 \,nm$ is greater than fringe width $\left(\beta_{2}\right)$ for $\lambda_{2}=$ $750\, nm$. This means that at though the central maxima of the two coincide, but first maximum for $\lambda_{1}=900$ $nm$ will be further away from the first maxima for $\lambda_{2}=750 \,nm$, and so on. A stage may come when this mismatch equals $\beta_{2}$, then again maxima of $\lambda_{1}=900$ $nm$, will coincide with a maxima of $\lambda_{2}=750\, nm$ let this correspond to nth order fringe for $\lambda_{1}$. Then it will correspond to $(n+1)^{t h}$ order fringe for $\lambda_{2}$.
Therefore $\frac{n \lambda_{1} D}{d}=\frac{(n+1) \lambda_{2} D}{d}$
$\Rightarrow n \times 900 \times 10^{-9}=(n+1) 750 \times 10^{-9} \Rightarrow n=5$
Minimum distance from central maxima
$=\frac{n \lambda_{1} D}{d}=\frac{5 \times 900 \times 10^{-9} \times 2}{2 \times 10^{-3}}=4.5 \,mm$